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Teaching video
A teaching recording for this chapter has not been published yet. Start with the worked example below and explore the related animations where available.
Browse grades and teaching videos01 · Read and understand
What you will learn
- Evaluate a force field along a parameterized path.
- Use a dot product inside a line integral.
- Check path orientation and a potential function.
Before you start
Vectors, dot products, polynomial integration, and basic partial derivatives.
Keep paper nearby. Read the question once for the context, then again to identify what is known and what you need to find.
Start with a question
A two-dimensional force field is F(x, y) = (2x, y) newtons. Along r(t) = (t, 2t) metres for 0 ≤ t ≤ 1, how much work is done?
Why this math matters
Only the force component along motion contributes to work. When force changes with position, multiply force by each small directed displacement and accumulate along the path. A vector line integral expresses this precisely. Some fields also have a potential, allowing an endpoint calculation that independently checks the integral.
Set up the model
A useful answer starts with clear assumptions:
- The coefficients have the units required to turn metre coordinates into newtons.
- t is a dimensionless path parameter increasing from zero to one.
- The ideal field is defined throughout the plane; work is measured in joules.
02 · Work through the example
Follow the reasoning, one step at a time.
Try to predict the next step before reading it. After each calculation, explain why the operation makes sense and how it helps answer the original question.
Work through it with Amar
See this example unfold.
The complete worked example, one idea at a time.
How do changing forces add up along a path?
PausedQuestion: Start with the question. Paused.
Question
Start with the question
A two-dimensional force field is F(x, y) = (2x, y) newtons. Along r(t) = (t, 2t) metres for 0 ≤ t ≤ 1, how much work is done?
Before you calculate
Read what is known and what you need to find. Make a prediction before moving to the first calculation.
Starts paused. Play advances through the full text at a reading pace; pause whenever you need more time. Previous, Next, and the phase buttons let you set your own pace. Playback pauses when this walkthrough leaves the screen or you switch tabs.
Your device’s reduced-motion setting keeps each phase still. Manual controls remain available. The full written solution stays below.
Describe force along the motion
r′(t) = (1, 2); F(r(t)) = (2t, 2t)
Substitute both path coordinates into the field. The derivative of the path supplies the directed displacement per unit parameter.
Accumulate the tangential contribution
W = ∫₀¹ F(r(t)) · r′(t) dt = ∫₀¹6t dt = 3 J
The dot product is 2t(1) + 2t(2). Integrating combines the changing contribution along the entire route.
Check using a potential
φ(x, y) = x² + y²/2; ∇φ = F; φ(1, 2) − φ(0, 0) = 3
Because the field is the gradient of this potential throughout the plane, its work depends only on endpoints for suitable paths.
The result
The force does 3 joules of work along the directed path.
Reversing the path reverses the sign. Endpoint-only calculation is special to conservative fields; it cannot be assumed for every force field. The potential used here is a mathematical work potential, with its stated gradient sign convention.
Common mistakes to catch
- Integrating force magnitude times distance discards the direction-dependent dot product.
- Not every vector field has a potential or path-independent work.
03 · Practice independently
Try it before revealing the answer.
Use paper or a calculator as needed. Write your units and reasoning, then open the hint or explanation to check your approach.
Practice 1
What work does the same field do when the path is traversed backward?
Show a hint
Reverse the endpoint subtraction.
Reveal answer and explanation
−3 joules
φ(0, 0) − φ(1, 2) = −3. Directed work changes sign under reversal.
Practice 2
For G = (−y, x), compute circulation around the unit circle counterclockwise.
Show a hint
Use r(θ) = (cos θ, sin θ), with 0 ≤ θ ≤ 2π in radians.
Reveal answer and explanation
2π
G(r) = (−sin θ, cos θ) equals r′, so their dot product is one and its integral is 2π.
Take the idea with you
A line integral combines a field, a path, and an orientation. Specify all three before calculating.
04 · Reflect and continue
Can you explain it in your own words?
Before moving on, explain the main idea without looking at the worked example. Try both practice questions, check your reasoning, and name one mistake you now know how to avoid. Return to a step if you still need support.
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